MathDB

3

Part of 2010 Romania National Olympiad

Problems(6)

4AM=3AC in regular pyramid VABCD

Source: Romanian MO 2010 Grade 8

8/6/2012
Let VABCDVABCD be a regular pyramid, having the square base ABCDABCD. Suppose that on the line ACAC lies a point MM such that VM=MBVM=MB and (VMB)⊥(VAB)(VMB)\perp (VAB). Prove that 4AM=3AC4AM=3AC.
Mircea Fianu
geometry3D geometrypyramidPythagorean Theoremgeometry proposed
No sequence of steps leaving 2011 blue squares

Source: Romanian MO 2010 Grade 7

8/6/2012
Each of the small squares of a 50×5050\times 50 table is coloured in red or blue. Initially all squares are red. A step means changing the colour of all squares on a row or on a column. a) Prove that there exists no sequence of steps, such that at the end there are exactly 20112011 blue squares. b) Describe a sequence of steps, such that at the end exactly 20102010 squares are blue.
Adriana & Lucian Dragomir
combinatorics proposedcombinatorics
Sets of solutions to floor equation

Source: Romanian MO 2010 Grade 9

8/6/2012
For any integer n≥2n\ge 2 denote by AnA_n the set of solutions of the equation x=⌊x2⌋+⌊x3⌋+⋯+⌊xn⌋.x=\left\lfloor\frac{x}{2}\right\rfloor+\left\lfloor\frac{x}{3}\right\rfloor+\cdots+\left\lfloor\frac{x}{n}\right\rfloor . a) Determine the set A2∪A3A_2\cup A_3. b) Prove that the set A=⋃n≥2AnA=\bigcup_{n\ge 2}A_n is finite and find max⁡A\max A.
Dan Nedeianu & Mihai Baluna
floor functionalgebra unsolvedalgebra
Arrangements into 10 groups with least number of triangles

Source: Romanian MO 2010 Grade 10

8/6/2012
In the plane are given 100100 points, such that no three of them are on the same line. The points are arranged in 1010 groups, any group containing at least 33 points. Any two points in the same group are joined by a segment. a) Determine which of the possible arrangements in 1010 such groups is the one giving the minimal numbers of triangles. b) Prove that there exists an arrangement in such groups where each segment can be coloured with one of three given colours and no triangle has all edges of the same colour.
Vasile Pop
Ramsey Theorycombinatorics unsolvedcombinatorics
Romania National Olympiad 2010 - Grade XI

Source:

4/10/2011
Let f:R→[0,∞)f:\mathbb{R}\rightarrow [0,\infty). Prove that f(x+y)≥(y+1)f(x), (∀)x∈Rf(x+y)\ge (y+1)f(x),\ (\forall)x\in \mathbb{R} if and only if the function g:R→[0,∞), g(x)=e−xf(x), (∀)x∈Rg:\mathbb{R}\rightarrow [0,\infty),\ g(x)=e^{-x}f(x),\ (\forall)x\in \mathbb{R} is increasing.
functionreal analysisreal analysis unsolved
Inequalities with |H|

Source: Romanian MO 2010 Grade 12

8/6/2012
Let GG be a finite group of order nn. Define the set H={x:x∈G and x2=e},H=\{x:x\in G\text{ and }x^2=e\}, where ee is the neutral element of GG. Let p=∣H∣p=|H| be the cardinality of HH. Prove that a) ∣H∩xH∣≥2p−n|H\cap xH|\ge 2p-n, for any x∈Gx\in G, where xH={xh:h∈H}xH=\{xh:h\in H\}. b) If p>3n4p>\frac{3n}{4}, then GG is commutative. c) If n2<p≤3n4\frac{n}{2}<p\le\frac{3n}{4}, then GG is non-commutative.
Marian Andronache
inequalitiesgroup theoryabstract algebrasuperior algebrasuperior algebra unsolved